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UY1: Sample questions for centre of mass (Set 1)

UY1 practice set on centre of mass with short hints and worked answers, including particle and lamina examples.

  • University Physics Year 1
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Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Cluster role + pathway links

This page is the archived bridge question set in the UY1 mechanics pathway.

1) At a glance

  • Prerequisite: weighted averages, basic CM formulas, symmetry.
  • Outcomes: choose a correct model quickly and solve CM questions with sign/units checks.
  • Key result: for composite bodies, treat removed parts as negative mass.
Common traps (composite bodies)
  • For cut-outs/holes, use a negative mass and keep the position sign (left/right, up/down).
  • Don’t cancel masses too early: keep ∑ mᵢ and ∑ mᵢxᵢ symbolic, then simplify.
  • Check a limiting case: hole size → 0 should give the full-body CM.

2) Setup (model, assumptions, coordinates/signs)

  • Choose an origin and axes before writing equations.
  • Use one consistent unit system.
  • For composite bodies:
x_CM = (∑ᵢ mᵢ xᵢ)/(∑ᵢ mᵢ)

with mᵢ < 0 for cut-out sections.

  • For uniform laminae, mass is proportional to area.

3) Core method / derivation

General problem-solving pipeline:

  1. Break body into simple known pieces.
  2. Assign each piece a signed mass (positive for present material, negative for holes).
  3. Write CM equations for x (and y if needed).
  4. Check signs and limits (if hole size → 0, result should return to full-body CM).

Quick checks:

  • Units: final CM coordinate must be in length units.
  • Sign: if mass removed on the right, CM should shift left.
  • Limit: tiny cut-out gives tiny shift.

4) Worked example

Two point masses: 2 kg at x = 0 and 6 kg at x = 4 m.

x_CM = (2(0) + 6(4))/(2 + 6) = 3 m

CM is closer to the heavier mass, as expected.

5) Practice set (hints + answers)

  1. A circular pizza of radius R has a circular hole of radius R/4 removed from the right side (hole center on the horizontal diameter, tangent internally).

    A uniform disc of radius R centred at the origin O on the x-axis, with a circular hole of radius R/4 removed on the right. The hole's centre is on the x-axis at 3R/4 from O, so the hole touches the disc's right edge from the inside.
    The hole touches the disc's right edge from the inside, so its centre lies 3R/4 to the right of O.

    Hint: treat hole as negative mass at x = +3R/4, full disk CM at x = 0. Answer:

    x_CM = (M(0)-(M/16)(3R/4))/(M-M/16) = -R/20

    So CM is shifted left by R/20.

  2. Equal masses at (0,0), (4,0), (0,2). Find CM. Hint: average coordinates. Answer: (x_CM,y_CM) = (4/3,2/3).

  3. Uniform rod from x = 0 to x = L has extra point mass m at x = L. Rod mass is 2m. Find CM. Hint: rod CM at L/2. Answer:

x_CM = ((2m)(L/2) + mL)/3m = 2L/3

6) Summary + next steps

  • Composite-body CM problems are bookkeeping with signed masses.
  • The fastest error checks are units, sign, and limiting behavior.
  • If symmetry exists, exploit it first.

Next steps: